Clicker Questions

1. The reason to take random samples is:¹

1. to make cause and effect conclusions
2. to get as many variables as possible
3. it’s easier to collect a large dataset
4. so that the data are a good representation of the population
5. I have no idea why one would take a random sample

2. The reason to allocate/assign explanatory variables is:²

1. to make cause and effect conclusions
2. to get as many variables as possible
3. it’s easier to collect a large dataset
4. so that the data are a good representation of the population
5. I have no idea what you mean by “allocate/assign” (or “explanatory variable” for that matter)

3. Approximately how big is a tweet?³
   1. 0.01Kb
   2. 0.1Kb
   3. 1Kb
   4. 100Kb
   5. 1000Kb = 1Mb

4. \(R^2\) measures:⁴

1. the proportion of variability in vote margin as explained by tweet share.
2. the proportion of variability in tweet share as explained by vote margin.
3. how appropriate the linear part of the linear model is.
4. whether or not particular variables should be included in the model.

5. R / R Studio / Quarto⁵
   1. all good
   2. started, progress is slow and steady
   3. started, very stuck
   4. haven’t started yet
   5. what do you mean by “R”?

6. Git / GitHub⁶
   1. all good
   2. started, progress is slow and steady
   3. started, very stuck
   4. haven’t started yet
   5. what do you mean by “Git”?

7. Which of the following includes talking to the remote version of GitHub?⁷
   1. changing your name (updating the YAML)
   2. committing the file(s)
   3. pushing the file(s)
   4. some of the above
   5. all of the above

8. What is the error?⁸
   1. poor assignment operator
   2. unmatched quotes
   3. improper syntax for function argument
   4. invalid object name
   5. no mistake

shup2 <-- "Hello to you!"

9. What is the error?⁹
   1. poor assignment operator
   2. unmatched quotes
   3. improper syntax for function argument
   4. invalid object name
   5. no mistake

3shup <-  "Hello to you!"

10. What is the error?¹⁰
    1. poor assignment operator
    2. unmatched quotes
    3. improper syntax for function argument
    4. invalid object name
    5. no mistake

shup4 <-  "Hello to you!

11. What is the error?¹¹
    1. poor assignment operator
    2. unmatched quotes
    3. improper syntax for function argument
    4. invalid object name
    5. no mistake

shup5 <-  date()

12. What is the error?¹²
    1. poor assignment operator
    2. unmatched quotes
    3. improper syntax for function argument
    4. invalid object name
    5. no mistake

shup6 <-  sqrt 10

13. Do you keep a calendar / schedule / planner?¹³
    1. Yes
    2. No

14. Do you keep a calendar / schedule / planner? If you answered “Yes” …¹⁴
    1. Yes, on Google Calendar
    2. Yes, on Calendar for macOS
    3. Yes, on Outlook for Windows
    4. Yes, in some other app
    5. Yes, by hand

15. Where should I put things I’ve created for the HW (e.g., data, .ics file, etc.)¹⁵
    1. Upload into remote GitHub directory
    2. In the local folder which also has the R project
    3. In my Downloads
    4. Somewhere on my Desktop
    5. In my Home directory

16. The goal of making a figure is…¹⁶
    1. To draw attention to your work.
    2. To facilitate comparisons.
    3. To provide as much information as possible.

17. A good reason to make a particular choice of a graph is:¹⁷
    1. Because the journal / field has particular expectations for how the data are presented.
    2. Because some variables naturally fit better on some graphs (e.g., numbers on scatter plots).
    3. Because that graphic displays the message you want as optimally as possible.

18. Why are the points orange?¹⁸
    1. R translates “navy” into orange.
    2. color must be specified in geom_point()
    3. color must be specified outside the aes() function
    4. the default plot color is orange

ggplot(data = Births78, 
       aes(x = date, y = births, color = "navy")) + 
  geom_point() +          
  labs(title = "US Births in 1978")

19. Why are the dots blue and the lines colored?¹⁹
    1. dot color is given as “navy”, line color is given as wday.
    2. both colors are specified in the ggplot() function.
    3. dot coloring takes precedence over line coloring.
    4. line coloring takes precedence over dot coloring.

20. Setting vs. Mapping. If I want information to be passed to all data points (not variable):²⁰
    1. map the information inside the aes() function.
    2. set the information outside the aes() function

21. The Snow figure was most successful at:²¹
    1. making the data stand out
    2. facilitating comparison
    3. putting the work in context
    4. simplifying the story

22. The Challenger figure(s) was(were) least successful at:²²
    1. making the data stand out
    2. facilitating comparison
    3. putting the work in context
    4. simplifying the story

23. The biggest difference between Snow and the Challenger was:²³
    1. The amount of information portrayed.
    2. One was better at displaying cause.
    3. One showed the relevant comparison better.
    4. One was more artistic.

24. Caffeine and Calories. What was the biggest concern over the average value axes?²⁴
    1. It isn’t at the origin.
    2. They should have used all the data possible to find averages.
    3. There wasn’t a random sample.
    4. There wasn’t a label explaining why the axes were where they were.

25. What is wrong with the following code?²⁵
    1. should only be one =
    2. Sydney should be lower case
    3. name should not be in quotes
    4. use mutate instead of filter
    5. babynames in wrong place

Result <- |> filter(babynames,
        name== "Sydney")

26. Which data represents the ideal format for ggplot2 and dplyr?²⁶

27. Each of the statements except one will accomplish the same calculation. Which one does not match?²⁷

#(a) 
babynames |> 
  group_by(year, sex) |> 
  summarize(totalBirths = sum(num))

#(b) 
group_by(babynames, year, sex) |> 
  summarize(totalBirths = sum(num))

#(c)
group_by(babynames, year, sex) |> 
  summarize(totalBirths = mean(num))

#(d)
temp <- group_by(babynames, year, sex)

summarize(temp, totalBirths = sum(num))

#(e)
summarize(group_by(babynames, year, sex), 
          totalBirths = sum(num))

28. Fill in Q1.²⁸
    1. filter()
    2. arrange()
    3. select()
    4. mutate()
    5. group_by()

result <- babynames |>
  Q1(name %in% c("Jane", "Mary")) |> 
  # just the Janes and Marys
  group_by(Q2, Q2) |> 
  summarize(total = Q3)

29. Fill in Q2.²⁹
    1. (year, sex)
    2. (year, name)
    3. (year, num)
    4. (sex, name)
    5. (sex, num)

result <- babynames |>
  Q1(name %in% c("Jane", "Mary")) |> 
  group_by(Q2, Q2) |> 
  # for each year for each name
  summarize(total = Q3)

30. Fill in Q3.³⁰
    1. n_distinct(name)
    2. n_distinct(n)
    3. sum(name)
    4. sum(num)
    5. mean(num)

result <- babynames |>
  Q1(name %in% c("Jane", "Mary")) |> 
  group_by(Q2, Q2) |> 
  summarize(total = Q3)
  # number of babies (each year, each name)

31. Running the code.³¹

babynames <- babynames::babynames |> 
  rename(num = n)

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  # just the Janes and Marys
  group_by(name, year) |> 
  # for each year for each name
  summarize(total = sum(num))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year total
   <chr> <dbl> <int>
 1 Jane   1880   215
 2 Jane   1881   216
 3 Jane   1882   254
 4 Jane   1883   247
 5 Jane   1884   295
 6 Jane   1885   330
 7 Jane   1886   306
 8 Jane   1887   288
 9 Jane   1888   446
10 Jane   1889   374
# ℹ 266 more rows

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(number = sum(num))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year number
   <chr> <dbl>  <int>
 1 Jane   1880    215
 2 Jane   1881    216
 3 Jane   1882    254
 4 Jane   1883    247
 5 Jane   1884    295
 6 Jane   1885    330
 7 Jane   1886    306
 8 Jane   1887    288
 9 Jane   1888    446
10 Jane   1889    374
# ℹ 266 more rows

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(n_distinct(name))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year `n_distinct(name)`
   <chr> <dbl>              <int>
 1 Jane   1880                  1
 2 Jane   1881                  1
 3 Jane   1882                  1
 4 Jane   1883                  1
 5 Jane   1884                  1
 6 Jane   1885                  1
 7 Jane   1886                  1
 8 Jane   1887                  1
 9 Jane   1888                  1
10 Jane   1889                  1
# ℹ 266 more rows

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(n_distinct(num))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year `n_distinct(num)`
   <chr> <dbl>             <int>
 1 Jane   1880                 1
 2 Jane   1881                 1
 3 Jane   1882                 1
 4 Jane   1883                 1
 5 Jane   1884                 1
 6 Jane   1885                 1
 7 Jane   1886                 1
 8 Jane   1887                 1
 9 Jane   1888                 1
10 Jane   1889                 1
# ℹ 266 more rows

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(sum(name))

Error in `summarize()`:
ℹ In argument: `sum(name)`.
ℹ In group 1: `name = "Jane"` and `year = 1880`.
Caused by error in `base::sum()`:
! invalid 'type' (character) of argument

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(mean(num))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year `mean(num)`
   <chr> <dbl>       <dbl>
 1 Jane   1880         215
 2 Jane   1881         216
 3 Jane   1882         254
 4 Jane   1883         247
 5 Jane   1884         295
 6 Jane   1885         330
 7 Jane   1886         306
 8 Jane   1887         288
 9 Jane   1888         446
10 Jane   1889         374
# ℹ 266 more rows

babynames |>
  filter(name %in% c("Jane", "Mary")) |> 
  group_by(name, year) |> 
  summarize(median(num))

# A tibble: 276 × 3
# Groups:   name [2]
   name   year `median(num)`
   <chr> <dbl>         <dbl>
 1 Jane   1880           215
 2 Jane   1881           216
 3 Jane   1882           254
 4 Jane   1883           247
 5 Jane   1884           295
 6 Jane   1885           330
 7 Jane   1886           306
 8 Jane   1887           288
 9 Jane   1888           446
10 Jane   1889           374
# ℹ 266 more rows

32. Fill in Q1.³²
    1. gdp
    2. year
    3. gdpval
    4. country
    5. –country

GDP |>  
  select(country = starts_with("Income"), everything()) |> 
       pivot_longer(cols = Q1, 
                    names_to = Q2, 
                    values_to = Q3)

33. Fill in Q2.³³
    1. gdp
    2. year
    3. gdpval
    4. country
    5. –country

GDP |>  
  select(country = starts_with("Income"), everything()) |> 
       pivot_longer(cols = Q1, 
                    names_to = Q2, 
                    values_to = Q3)

34. Fill in Q3.³⁴
    1. gdp
    2. year
    3. gdpval
    4. country
    5. –country

GDP |>  
  select(country = starts_with("Income"), everything()) |> 
       pivot_longer(cols = Q1, 
                    names_to = Q2, 
                    values_to = Q3)

35. Response to stimulus (in ms) after only 3 hrs of sleep for 9 days. You want to make a plot with the subject’s reaction time (y-axis) vs the number of days of sleep restriction (x-axis) using the following ggplot() code. Which data frame should you use?³⁵
    1. use raw data
    2. use pivot_wider() on raw data
    3. use pivot_longer() on raw data

ggplot(___, aes(x = ___, y = ___, color = ___)) + 
  geom_line()

# A tibble: 18 × 11
   Subject day_0 day_1 day_2 day_3 day_4 day_5 day_6 day_7 day_8 day_9
     <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 1     308  250.  259.  251.  321.  357.  415.  382.  290.  431.  466.
 2     309  223.  205.  203.  205.  208.  216.  214.  218.  224.  237.
 3     310  199.  194.  234.  233.  229.  220.  235.  256.  261.  248.
 4     330  322.  300.  284.  285.  286.  298.  280.  318.  305.  354.
 5     331  288.  285   302.  320.  316.  293.  290.  335.  294.  372.
 6     332  235.  243.  273.  310.  317.  310   454.  347.  330.  254.
 7     333  284.  290.  277.  300.  297.  338.  332.  349.  333.  362.
 8     334  265.  276.  243.  255.  279.  284.  306.  332.  336.  377.
 9     335  242.  274.  254.  271.  251.  255.  245.  235.  236.  237.
10     337  312.  314.  292.  346.  366.  392.  404.  417.  456.  459.
11     349  236.  230.  239.  255.  251.  270.  282.  308.  336.  352.
12     350  256.  243.  256.  256.  269.  330.  379.  363.  394.  389.
13     351  251.  300.  270.  281.  272.  305.  288.  267.  322.  348.
14     352  222.  298.  327.  347.  349.  353.  354.  360.  376.  389.
15     369  272.  268.  257.  278.  315.  317.  298.  348.  340.  367.
16     370  225.  235.  239.  240.  268.  344.  281.  348.  365.  372.
17     371  270.  272.  278.  282.  279.  285.  259.  305.  351.  369.
18     372  269.  273.  298.  311.  287.  330.  334.  343.  369.  364.

36. Consider band members from the Beatles and the Rolling Stones. Who is removed in a right_join()?³⁶

1. Mick
2. John
3. Paul
4. Keith
5. Impossible to know

band_members |> 
  right_join(band_instruments, by = "name")

band_members

# A tibble: 3 × 2
  name  band   
  <chr> <chr>  
1 Mick  Stones 
2 John  Beatles
3 Paul  Beatles

band_instruments

# A tibble: 3 × 2
  name  plays 
  <chr> <chr> 
1 John  guitar
2 Paul  bass  
3 Keith guitar

37. Consider band members from the Beatles and the Rolling Stones. Which variables are removed in a right_join()?³⁷

1. name
2. band
3. plays
4. none of them

band_members |> 
  right_join(band_instruments, by = "name")

band_members

# A tibble: 3 × 2
  name  band   
  <chr> <chr>  
1 Mick  Stones 
2 John  Beatles
3 Paul  Beatles

band_instruments

# A tibble: 3 × 2
  name  plays 
  <chr> <chr> 
1 John  guitar
2 Paul  bass  
3 Keith guitar

38. What happens to Mick’s plays variable in a full_join()?³⁸

1. Mick is removed
2. changes to guitar
3. changes to bass
4. NA
5. NULL

band_members |> 
  full_join(band_instruments, by = "name")

band_members

# A tibble: 3 × 2
  name  band   
  <chr> <chr>  
1 Mick  Stones 
2 John  Beatles
3 Paul  Beatles

band_instruments

# A tibble: 3 × 2
  name  plays 
  <chr> <chr> 
1 John  guitar
2 Paul  bass  
3 Keith guitar

39. Consider the addTen() function. The following output is a result of which map_*() call?³⁹

1. map(c(1,4,7), addTen)
2. map_dbl(c(1,4,7), addTen)
3. map_chr(c(1,4,7), addTen)
4. map_lgl(c(1,4,7), addTen)

addTen <- function(wow) {
  return(wow + 10)
}

[1] "11.000000" "14.000000" "17.000000"

40. Which of the following input is allowed?⁴⁰
    1. map(c(1, 4, 7), addTen)
    2. map(list(1, 4, 7), addTen)
    3. map(data.frame(a=1, b=4, c=7), addTen)
    4. some of the above
    5. all of the above

41. Which of the following produces a different output?⁴¹
    1. map(c(1, 4, 7), addTen)
    2. map(c(1, 4, 7), ~addTen(.x))
    3. map(c(1, 4, 7), ~addTen)
    4. map(c(1, 4, 7), function(hi) (hi + 10))
    5. map(c(1, 4, 7), ~(.x + 10))

42. What will the following code output?⁴²
    1. 3 random normals
    2. 6 random normals
    3. 18 random normals

input

# A tibble: 3 × 3
      n  mean    sd
  <dbl> <dbl> <dbl>
1     1     1     3
2     2     3     1
3     3    47    10

input |> 
  pmap(rnorm)

43. In R the ifelse() function takes the arguments:⁴³

1. question, yes, no
2. question, no, yes
3. statement, yes, no
4. statement, no, yes
5. option1, option2, option3

44. What is the output of the following:⁴⁴
    1. “cat”, 30, “cat”, “cat”, 6
    2. “cat”, “30”, “cat”, “cat”, “6”
    3. 1, “cat”, 5, “cat”, “cat”
    4. 1, “cat”, 5, NA, “cat”
    5. “1”, “cat”, “5”, NA, “cat”

data <- c(1, 30, 5, NA, 6)

ifelse(data > 5, "cat", data)

45. In R, the set.seed() function⁴⁵

1. makes your computations go faster
2. keeps track of your computation time
3. provides an important parameter
4. repeats the function
5. makes your results reproducible

46. If I run a hypothesis test with a type I error cut off of \(\alpha = 0.05\) and the null hypothesis is true, what is the probability of rejecting \(H_0\)?⁴⁶

1. 0.01
2. 0.05
3. 0.1
4. I don’t know.
5. No one knows.

47. If I run a hypothesis test with a type I error cut off of \(\alpha = 0.05\) and the null hypothesis is true, and also the technical conditions do not hold what is the probability of rejecting \(H_0\)?⁴⁷

1. 0.01
2. 0.05
3. 0.1
4. I don’t know.
5. No one knows.

48. If I run a hypothesis test with a type I error cut off of \(\alpha = 0.05\) and the null hypothesis is false, what is the probability of rejecting \(H_0\)?⁴⁸

1. 0.01
2. 0.05
3. 0.1
4. I don’t know.
5. No one knows.

49. If I aim to create a 95% confidence interval, and the technical conditions hold, what is the probability that the CI will contain the true value of the parameter?⁴⁹

1. 0.90
2. 0.95
3. 0.99
4. I don’t know.
5. No one knows.

50. If I aim to create a 95% confidence interval, and the technical conditions do not hold, what is the probability that the CI will contain the true value of the parameter?⁵⁰

1. 0.90
2. 0.95
3. 0.99
4. I don’t know.
5. No one knows.

51. We typically compare means (across two groups) instead of medians because:⁵¹

1. we don’t know the SE of the difference of medians
2. means are inherently more interesting than medians
3. permutation tests don’t work with medians
4. the Central Limit Theorem doesn’t apply for medians.

52. What are the technical assumptions for a t-test?⁵²

1. none
2. normal data
3. \(n \geq 30\)
4. random sampling / random allocation for appropriate conclusions

53. What are the technical conditions for permutation tests?⁵³

1. none
2. normal data
3. \(n \geq 30\)
4. random sampling / random allocation for appropriate conclusions

54. Follow up to permutation test: the assumptions change based on whether the statistic used is the mean, median, proportion, etc.⁵⁴

1. TRUE
2. FALSE

55. Why care about the distribution of the test statistic?⁵⁵

1. Better estimator
2. So we can find rejection region
3. So we can control power
4. Because we love the CLT

56. Given statistic T = r(X), how do we find a (sensible) test?⁵⁶

1. Maximize power
2. Minimize type I error
3. Control type I error
4. Minimize type II error
5. Control type II error

57. Type I error is⁵⁷

1. We give him a raise when he deserves it.
2. We don’t give him a raise when he deserves it.
3. We give him a raise when he doesn’t deserve it.
4. We don’t give him a raise when he doesn’t deserve it.

58. Type II error is⁵⁸

1. We give him a raise when he deserves it.
2. We don’t give him a raise when he deserves it.
3. We give him a raise when he doesn’t deserve it.
4. We don’t give him a raise when he doesn’t deserve it.

59. Power is the probability that:⁵⁹

1. We give him a raise when he deserves it.
2. We don’t give him a raise when he deserves it.
3. We give him a raise when he doesn’t deserve it.
4. We don’t give him a raise when he doesn’t deserve it.

60. Why don’t we always reject \(H_0\)?⁶⁰
    1. type I error too high
    2. type II error too high
    3. level of sig too high
    4. power too high

61. The player is more worried about⁶¹
    1. A type I error
    2. A type II error

62. The coach is more worried about⁶²
    1. A type I error
    2. A type II error

63. Increasing your sample size⁶³
    1. Increases your power
    2. Decreases your power

64. Making your significance level more stringent (\(\alpha\) smaller)⁶⁴

1. Increases your power
2. Decreases your power

65. A more extreme alternative⁶⁵
    1. Increases your power
    2. Decreases your power

66. What is the primary reason to use a permutation test (instead of a test built on calculus)?⁶⁶

1. more power
2. lower type I error
3. more resistant to outliers
4. can be done on statistics with unknown sampling distributions

67. What is the primary reason to bootstrap a CI (instead of creating a CI from calculus)?⁶⁷

1. larger coverage probabilities
2. narrower intervals
3. more resistant to outliers
4. can be done on statistics with unknown sampling distributions

68. Which of the following could not possibly be a bootstrap sample from the vector: c(4, 10, 8, 1, 2, 4)⁶⁸
    1. c(4, 4, 4, 4, 4, 4)
    2. c(4, 10, 8, 1, 2, 4)
    3. c(1, 2, 2, 4, 4, 2)
    4. c(10, 8, 1, 1, 8, 10)
    5. c(1, 2, 4, 3, 4, 10)

69. You have a sample of size n = 50. You sample with replacement 1000 times to get 1000 bootstrap samples. What is the sample size of each bootstrap sample?⁶⁹

1. 50
2. 1000

70. You have a sample of size n = 50. You sample with replacement 1000 times to get 1000 bootstrap samples. How many bootstrap statistics will you have?⁷⁰

1. 50
2. 1000

71. The bootstrap distribution is centered around the⁷¹

1. population parameter
2. sample statistic
3. bootstrap statistic
4. bootstrap parameter

72. 95% CI for the difference in proportions:⁷²
    1. (0.15, 0.173)
    2. (0.025, 0.975)
    3. (0.72, 0.87)
    4. (0.70, 0.873)
    5. (0.12, 0.179)

73. Suppose a 95% bootstrap CI for the difference in trimmed means was (3,9), would you reject H0?⁷³ (uh… What is the null hypothesis here???)

1. yes
2. no
3. not enough information to know

74. Given the situation where \(H_a\) is TRUE. Consider 100 CIs (for true difference in means, where each of the 100 CIs is created using a different dataset). The power of the test can be approximated by:⁷⁴

1. The proportion that contain the true difference in means.
2. The proportion that do not contain the true difference in means.
3. The proportion that contain zero.
4. The proportion that do not contain zero.

75. Small k in k-NN will help reduce the risk of overfitting.⁷⁵
    1. TRUE
    2. FALSE

76. The training error for 1-NN classifier is zero.⁷⁶
    1. TRUE
    2. FALSE

77. Generally, the k-NN algorithm can take any distance measure.⁷⁷

1. TRUE
2. FALSE

78. In R, the kknn method can use any distance measure.⁷⁸

1. TRUE
2. FALSE

79. The k in k-NN refers to⁷⁹
    1. k groups
    2. k partitions
    3. k neighbors

80. the V in V-fold CV refers to⁸⁰
    1. V groups
    2. V partitions
    3. V neighbors

81. All of the following are TRUE for the use of CART except for:⁸¹

1. Can deal with missing data
2. Require the assumptions of statistical models
3. Variable selection is automatic
4. Produce rules that are easy to interpret and implement

82. Simplifying the decision tree by pruning peripheral branches will cause over fitting.⁸²

1. TRUE
2. FALSE

83. All are true with regards to PRUNING except:⁸³

1. Multiple (sequential) trees are possible to create by pruning
2. CART lets tree grow to full extent, then prunes it
   
3. Pruning generates successively smaller trees by pruning leaves
4. Pruning is only beneficial when purity improvement is statistically significant

84. Regression trees are invariant to monotonic transformations of:⁸⁴

1. the explanatory (predictor) variables
2. the response variable
3. both types of variables
4. none of the variables

85. CART suffers from⁸⁵
    1. high variance
    2. high bias

86. bagging uses bootstrapping on:⁸⁶
    1. the variables
    2. the observations
    3. both
    4. neither

87. oob samples⁸⁷

1. are in the test data
2. are in the training data and provide independent predictions
3. are in the training data but do not provide independent predictions

88. oob samples are great because⁸⁸

1. oob is “boo” spelled backwards
2. oob samples allow for independent predictions
3. oob samples allow for more predictions than a “test group”
4. oob data frame is always bigger than the test sample data frame
5. some of the above

89. bagging is random forests with:⁸⁹

1. m = # predictor variables
2. all the observations
3. the most important predictor variables isolated
4. cross validation to choose m

90. We have 80 training observations and 20 test observations. To get the test MSE, we need⁹⁰

1. 20 predictions from all trees
2. 20 predictions from oob trees
3. 80 predictions form all trees
4. 80 predictions from oob trees

91. With random forests, the value for m is chosen⁹¹

1. using OOB error rate
2. as p/3
3. as sqrt(p)
4. using cross validation

92. A tuning parameter:⁹²

1. makes the model fit the training data as well as possible.
2. makes the model fit the test data as well as possible.
3. allows for a good model that does not overfit the data.

93. With binary response and X1 and X2 continuous, kNN (k=1) creates a linear decision boundary.⁹³

1. TRUE
2. FALSE

94. With binary response and X1 and X2 continuous, a classification tree with one split creates a linear decision boundary.⁹⁴

1. TRUE
2. FALSE

95. With binary response and X1 and X2 continuous, a classification tree with one split creates the best linear decision boundary.⁹⁵

1. TRUE
2. FALSE

96. If the data are linearly separable, there exists a “widest street”.⁹⁶

1. yes
2. no
3. up to a constant
4. with the alpha values “tuned” appropriately

97. In the case of linearly separable data, the SVM:⁹⁷

1. has a tuning parameter of \(\alpha\)
2. has a tuning parameter of dimension
3. has no tuning parameters

98. Linear models are similar to SVM with linearly separable data in that they optimize the model instead of tuning it.⁹⁸
    1. TRUE
    2. FALSE

99. If the data have a complex boundary, the value of gamma in RBF kernel should be:⁹⁹
    1. Big
    2. Small

100. If the data have a simple boundary, the value of gamma in RBF kernel should be:¹⁰⁰
     1. Big
     2. Small

101. If I am using all features of my dataset and I achieve 100% accuracy on my training set, but ~70% on validation set, what should I look out for?¹⁰¹
     1. Underfitting
     2. Nothing, the model is perfect
     3. Overfitting

102. For a large value of C, the model is expected to¹⁰²
     1. overfit the training data more as compared to a small C b. overfit the training data less as compared to a small C c. not related to overfitting the data

103. Suppose you have trained an SVM with linear decision boundary after training SVM, you correctly infer that your SVM model is under fitting.
     Which of the following option would make you more likely to consider iterating the SVM next time?¹⁰³:
     1. You want to increase your data points
     2. You want to decrease your data points
     3. You will try to calculate more variables
     4. You will try to reduce the features

102. Suppose you have trained an SVM with linear decision boundary after training SVM, you correctly infer that your SVM model is under fitting.

Suppose you gave the correct answer in previous question. What do you think that is actually happening?

1. We are lowering the bias
2. We are lowering the variance
3. We are increasing the bias
4. We are increasing the variance

1. 1 and 2
2. 2 and 3
3. 1 and 4
4. 2 and 4

103. Suppose you are using SVM with linear kernel of polynomial degree 2, Now think that you have applied this on data and found that it perfectly fit the data that means, Training and testing accuracy is 100%.

You increase the complexity(degree of polynomial of this kernel). What will happen?

1. Increasing the complexity will overfit the data
2. Increasing the complexity will underfit the data
3. Nothing will happen since your model was already 100% accurate
4. None of these

  104. Suppose you are using SVM with linear kernel of polynomial degree 2, Now think that you have applied this on data and found that it perfectly fit the data that means, Training and testing accuracy is 100%.

In the previous question after increasing the complexity you found that training accuracy was still 100%. According to you what is the reason behind that?

1. Since data are fixed and we are fitting more polynomial terms, the algorithm starts memorizing everything in the data
2. Since data are fixed, SVM doesn’t need to search in high dimensional space

1. 1
2. 2
3. 1 and 2
4. None of these

105. The cost parameter in the SVM means:

1. The number of cross-validations to be made
2. The kernel to be used
3. The tradeoff between misclassification and simplicity of the model
4. None of the above

  106. Suppose you have trained an SVM classifier with a Gaussian kernel, and it learned the following decision boundary on the training set:

You suspect that the SVM is underfitting your dataset. Should you try increasing or decreasing C? Increasing or decreasing gamma?

1. decreasing C and/or decrease gamma
2. decreasing C and/or increase gamma
3. increasing C and/or decrease gamma
4. increasing C and/or increase gamma  

107. Suppose you have trained an SVM classifier with a Gaussian kernel, and it learned the following decision boundary on the training set:

When you measure the SVM’s performance on a cross validation set, it does poorly. Should you try increasing or decreasing C? Increasing or decreasing gamma?

1. decreasing C and/or decrease gamma
2. decreasing C and/or increase gamma
3. increasing C and/or decrease gamma
4. increasing C and/or increase gamma  

108. Cross validation will guarantee that the model does not overfit.

1. TRUE

2. FALSE

109. The biggest problem with missing data is the resulting small sample size.

1. True

2. False

110. Which statement is not true about cluster analysis?

1. Objects in each cluster tend to be similar to each other and dissimilar to objects in the other clusters.

2. Cluster analysis is a type of unsupervised learning.

3. Groups or clusters are suggested by the data, not defined a priori.

4. Cluster analysis is a technique for analyzing data when the response variable is categorical and the predictor variables are continuous in nature.

111. A _____ or tree graph is a graphical device for displaying clustering results. Vertical lines represent clusters that are joined together. The position of the line on the scale indicates the distances at which clusters were joined.

1. dendrogram

2.   scatterplot

3.   scree plot

4.   segment plot 

112. _____ is a clustering procedure characterized by the development of a tree-like structure.

1. Partitioning clustering
   
2. Hierarchical clustering
3. Divisive clustering
4. Agglomerative clustering

113. _____ is a clustering procedure where all objects start out in one giant cluster.

Clusters are formed by dividing this cluster into smaller and smaller clusters.

1. Non-hierarchical clustering
   
2. Hierarchical clustering
3. Divisive clustering
4. Agglomerative clustering  

114. The _____ method uses information on all pairs of distances, not merely the minimum or maximum distances.

1. single linkage
2. medium linkage
3. complete linkage
4. average linkage

115. Hierarchical clustering is deterministic, but k-means clustering is not.

1. TRUE

2. FALSE

  116. k-means is a clustering procedure characterized referred to as ________.

1. Partitioning clustering
   
2. Hierarchical clustering
3. Divisive clustering
4. Agglomerative clustering

117. One method of assessing reliability and validity of clustering is to use different methods of clustering and compare the results.

1. TRUE

2. FALSE

118. The choice of k, the number of clusters to partition a set of data into,…

1. is a personal choice that shouldn’t be discussed in public
2. depends on why you are clustering the data
3. should always be as large as your computer system can handle
4. has maximum 10

119. Which of the following is required by k-means clustering?

1. defined distance metric
2. number of clusters
3. initial guess as to cluster centroids
4. all of the above
5. some of the above

120. For which of the following tasks might clustering be a suitable approach?

1. Given sales data from many products in a supermarket, estimate future sales for each of these products.
2. Given a database of information about your users, automatically group them into different market segments.
3. From the user’s usage patterns on a website, identify different user groups.
4. Given historical weather records, predict if tomorrow’s weather will be sunny or rainy.

121. k-means is an iterative algorithm, and two of the following steps are repeatedly carried out. Which two?
122. Assign each point to its nearest cluster
123. Test on the cross-validation set
124. Update the cluster centroids based the current assignment
125. Using the elbow method to choose K

1. 1 & 2
2. 1 & 3
3. 1 & 4
4. 2 & 3
5. 2 & 4
6. 3 & 4